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2 years ago

A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If

Physics

1 answer:

Darya [45]2 years ago

5 0

The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

<h3>How to determine the force</h3>

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1 and q2 = charges = 1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F = $9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }$

F = $9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}$

F = $9* 10^9 * 7. 14* 10^-21$

F = $6. 426 * 10^-11$ Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

Learn more about electrostatic force here:

brainly.com/question/8424563

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The diameter of a 12-gauge copper wire is 0.081 in. The maximum safe current it can 17) carry (in order to prevent fire danger i

Soloha48 [4]

Answer:

0.44m/s

Explanation:

drift velocity=I/nAq

diameter 12 gauge

wire=0.081inches=0.081*2.5=0.2025cm radius=0.10125cm area=pi*R^2 =20/8.5*10^22*3.14*0.10125^2*10^-4*1.6*10^-19*

V = 0.44m/s

6 0

4 years ago

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a

Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

$w_1g = 1.4*9.81*y*sin30$ = - 6.867y

Workdone by gravity on the mass of the suspended block is:

$w_2g = 0.06*9.81$ = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y² - 6.867y + 0.5886 = 0.0588

-20 y² - 6.867y + 0.5886 - 0.0588

-20 y² - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y² + 6.867y - 0.5298 = 0

Using the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where;

a = 20 ; b = 6.867 c= - 0.5298

$\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$

$= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$ = 0.0649 OR −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0

4 years ago

What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)

puteri [66]

Answer:

When we have a current I, we will have a magnetic field perpendicular to this current.

Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".

Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.

Of course, the intensity of the magnetic field is proportional to the current, given by:

B = (μ*I)/(2*π*r)

Where:

μ is a constant, I is the current and r is the distance between to the current.

Now remember that for a resistor:

R = ρ*L/A

R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.

If we increase the area of the wire (if we use a thicker wire).

And the relation between resistance and current is:

I = V/R

Where V is the voltaje.

Now, if we use a thicker wire, then the cross-section area of the wire increases.

Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.

And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.

If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.

3 0

3 years ago

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

vovangra [49]

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

Momentum = mass x velocity

Since the speed of the ball is unknown:

Velocity = $\frac{momentum }{mass}$

= $\frac{0.78}{0.15}$

= 5.2m/s

The speed of the baseball before it lands is 5.2m/s

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

4 0

3 years ago

Name the following compound: CH2 ≡ CH2

iragen [17]

I think its ethylene, tell me if i'm wrong

4 0

3 years ago